What is the slope of the line tangent to $f(x) = x^{2}+2x-4$ at $x = -3$ ?
Explanation: The slope of the tangent line is $ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ $ = \lim_{h \to 0} \frac{((x+h)^{2}+2(x+h)-4) - (x^{2}+2x-4)}{h}$ $ = \lim_{h \to 0} \frac{(x^{2}+2x h+h^{2}+2(x+h)-4) - (x^{2}+2x-4)}{h}$ $ = \lim_{h \to 0} \frac{x^{2}+2(x h)+h^{2}+2x+2h-4-x^{2}-2x+4}{h}$ $ = \lim_{h \to 0} \frac{2(x h)+h^{2}+2h}{h}$ $ = \lim_{h \to 0} 2x+h+2$ $ = 2x+2$ $ = (2)(-3)+2$ $ = -4$